Question

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Answer 1

Draw OM perpendicular to AB and ON perpendicular to CD 

Join OB and OC

BM = \(\frac{AB}{2}\) = \(\frac{5}{2}\)

(Perpendicular from centre bisects the chord)

ND = \(\frac{CD}{2}\) = \(\frac{11}{2}\)

Let, 

ON be r so OM will be (6 – x)

In ΔMOB,

OM² + MB² = OB²

(6 – x)² + (\(\frac{5}{2}\))² = OB²

36 + x² – 12x + \(\frac{25}{4}\) = OB²  ....(i)

In ΔNOD,

ON² + ND² = OD²

x² + (\(\frac{11}{2}\))² = OD²

x² + \(\frac{121}{4}\) = OD² ...(ii)

We have, 

OB = OD 

(Radii of same circle) 

So from (i) and (ii), we get

36 + x² + 2x +\(\frac{25}{4}\) = x² + \(\frac{121}{4}\)

12x = 36 + \(\frac{25}{4}\) - \(\frac{121}{4}\)